Find maximum of object field inside an ArrayList

Question

I have an ArrayList of Nodes. One of the fields of Node class is level.

I would like to find the level of the node that has the maximum level in the list.

This code gives me an NullPointerException...

private int findMaxLevel(ArrayList<Node> nodes)
{
    int level = 0;
    for(Node node : nodes)
    {
        if(node.getLevel() > level)
        {
            level = node.getLevel();
        }
    }
    return level;
}

I've tried several implementations of this page : Sorting an ArrayList of Contacts based on name? but I didn't found the solution.

EDIT :

Thank you for your suggestions, the level should never be null but I've forgot to add this.level = level in the Node class constructor....

Show the implementation of `Node` and a minimal complete code that gives the error. — Mad Physicist, May 09 '17 at 14:50
@JohnKane. It appears that `level` is an `int`, and therefore never `null`. The issue is most likely that there are `null` elements in the list. — Mad Physicist, May 09 '17 at 14:51
@ΦXocę웃Пepeúpaツ. Why in the world would you want to do that? Sorting will turn an O(n) opeation into an O(nlogn) — Mad Physicist, May 09 '17 at 14:52
I was assuming that it could be an Integer (since no code was shown) — John Kane, May 09 '17 at 14:52
You are missing the most important information: The JRE exactly tells you in which line of code the NPE occurs! Without that info, we can only guess what is the reason for NPE! — Gyro Gearless, May 09 '17 at 15:04

score 0 · Answer 1 · answered May 09 '17 at 14:55

What is the null reference? What does the exceptions says? Anyway, the only reference in this piece of code is nodes so I assume it is it (getLevel() returns int type which is not nullable) Maybe you are sending the ArrayList<Node> before creating it? Maybe you are using the function like this somewhere:

findMaxLevel(null);

or

ArrayList<Node> nodes = null; // Something like that
findMaxLevel(nodes);

ΦXocę 웃 Пepeúpa ツ · Answer 2 · 2017-05-09T15:03:08.663

0

Putting apart the NPE...

you can implement comparable in the Node class and use a Collections.max to get the max element in the list.

int max = Collections.max(a);
System.out.println(max);

but why overriding comparable:?

Collections.max() is implemented as:

public static <T extends Object & Comparable<? super T>> T max(Collection<? extends T> coll) {
    Iterator<? extends T> i = coll.iterator();
    T candidate = i.next();

    while (i.hasNext()) {
        T next = i.next();
        if (next.compareTo(candidate) > 0)
            candidate = next;
    }
    return candidate;
}

so compareTo will be called to find the max element

edited May 09 '17 at 15:03

answered May 09 '17 at 14:57

ΦXocę 웃 Пepeúpa ツ

47,427
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Nope. Overriding `equals` does not define an order. You want to implement `Comparable` to make `Collections.max()` work! – Gyro Gearless May 09 '17 at 15:00

Anton Balaniuc · Answer 3 · 2017-05-09T15:12:21.633

You might use stream api for that. I assume that your node class has an int field level

 class Node {
            private int level;
            public int getLevel() {
                return  level;
            }
            public void setLevel(int level) {
                this.level = level;
            }
 }

Now we need to find out maxim level in a List with Nodes:

Optional<Integer> maxLevel = nodes.stream().filter(Objects::nonNull)
                .map(Node::getLevel).max(Comparator.<Integer>naturalOrder());

I hope this helps.

score 0 · Answer 4 · answered May 09 '17 at 15:02

Keeping things simple with this one-liner:

    Node max = nodes.stream()
                    .filter(Objects::nonNull)
                    .max((Node node, Node t1) -> t1.level < node.level ? 1 : -1)
                    .get();

Make sure to check and see if the list is empty first or use Optional, otherwise you might get a NoSuchElementException.

4 Answers4